Ahlan wa Sahlan: Functional Modern Standard Arabic for by Mahdi Alosh

By Mahdi Alosh

This textbook is designed to hide the 1st yr of guideline in smooth common Arabic. observed by means of an instructor's handbook and an audio programme, it is going to educate scholars to learn, converse, and write Arabic. The textual content provides an enticing tale that contains Adnan, a Syrian scholar learning within the united states, and Michael, an American pupil learning in Cairo. In diaries, letters, and postcards, the 2 scholars describe their techniques and actions, revealing how a non-American perspectives American tradition and the way the Arabic tradition is skilled by means of an American scholar. The textual content additionally presents information regarding the geography of the Arab global, renowned characters in heritage, festivities in Arab tradition, the media, everyday life, and the kinfolk. routines in comprehension, vocabulary, grammar, and writing attend to either shape and which means and increase useful talents and information in regards to the Arabic sound, writing, and language platforms.

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Sample text

The final matrix corresponds to the linear system ⎧ ⎨ x − y + z = 20 y − 2z = 0 . ⎩ z = 31 The third equation in this system yields z = 31. Plugging this value into the second equation produces y − 2(31) = 0, or y = 62. Finally, we substitute these values into the first equation to get x − 62 + 31 = 20, or x = 51. Hence, there are 51 nickels, 62 dimes, and 31 quarters. (4) First, plug each of the three given linear system: ⎧ ⎨ 18 = 9 = ⎩ 13 = points into the equation y = ax2 + bx + c. This yields the following 9a + 3b + c ←− using (3, 18) 4a + 2b + c ←− using (2, 9) .

The last equation states that x5 = 2. Hence, the complete solution set is {(2b − d − 4, b, 2d + 5, d, 2) | b, d ∈ R}. 1 is ⎡ 5 ⎤ 1 − 12 − 74 4 ⎥ ⎢ 1 − 14 − 74 ⎥ ⎢ 0 ⎢ ⎥. ⎣ 0 0 1 3 ⎦ 0 0 0 0 We target the entries above the pivots to obtain the reduced row echelon form matrix. First, we use the (2,2) entry as the pivot and target the (1,2) entry. 2 Next, we use the (3,3) entry as a pivot and target the (1,3) and (2,3) entries. ⎡ ⎤ 1 0 0 6 ⎢ 0 1 0 1 ← ( 15 −1 ⎥ 8 )× 3 + 1 ⎢ ⎥ ⎣ 0 0 1 3 ⎦ 2 ← ( 14 ) × 3 + 2 0 0 0 0 Ignoring the last row, which gives the equation 0 = 0, this matrix yields the linear system ⎧ = 6 ⎨ x1 = −1 , x2 ⎩ 3 x3 = producing the unique solution (6, −1, 3).

Hence, the pivot moves to the (2,3) entry. We convert that to 1. ⎡ ⎤ 10 19 4 0 1 4 − 27 7 7 7 ⎢ ⎥ ⎢ 0 0 1 0 ⎥ 3 −15 −32 ⎢ ⎥ 2 ← (7) 2 ⎢ 0 0 6 12 20 2 0 ⎥ ⎣ ⎦ 7 7 7 7 10 26 0 0 0 − 37 − 97 7 7 Next, we target the (1,3), (3,3), and (4,3) ⎡ 1 4 1 ← (− 47 ) × 2 + 1 ⎢ ⎢ 0 0 3 ← (− 27 ) × 2 + 3 ⎣ 0 0 4 ← ( 37 ) × 2 + 4 0 0 entries. 0 −2 1 3 0 0 0 0 10 21 −15 −32 6 12 −5 −10 ⎤ 0 0 ⎥ ⎥ 0 ⎦ 0 We try to move the pivot to the third column but cannot due to the zeroes in the (3,4) and (4,4) entries. Hence, the pivot moves to the (3,5) entry.

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